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EDIT: $R_n$ is the right endpoint of the Riemann sum

Calculate $R_n$ for the function $$f(x) = -\frac{x^2}4-5$$ on interval [0,2]

What I've done so far

$ \Delta x = \frac 2n$ and $f(x_i^*) = \frac{2i}n $

Which leads to $$ \sum_{i=1}^n f(x_i^*)\ \Delta x$$

And once you put the numbers in

$$ \sum_{i=1}^n\left( -\frac{\left(\frac{2i}n\right)^2}4-5\right) \cdot \frac 2n $$

Which leads to my issue of expanding and getting an $i$ or $i^2$ alone.

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  • $\begingroup$ @T.Bongers Just added that in the question. $\endgroup$
    – impact_sv1
    Jan 17, 2016 at 21:43

1 Answer 1

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You have $$ \sum_{i=1}^n f(x_i^*)\ \Delta x=-\frac {2}{n^3} \sum_{i=1}^n i^2- \frac {10}{n}\sum_{i=1}^n 1 $$ then use $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ which is proved here to get, as $n \to \infty$, $$ \sum_{i=1}^n f(x_i^*)\ \Delta x \to -\frac{32}3 $$

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    $\begingroup$ Hey thanks dude! Appreciate that! $\endgroup$
    – impact_sv1
    Jan 17, 2016 at 22:03
  • $\begingroup$ You are welcome. @impact_sv1 $\endgroup$ Jan 17, 2016 at 22:06

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